The substitution 9 transforms a quadratic form Q xT AX sig

The substitution (9) transforms a quadratic form Q = x^T AX = sigma_j = 1^R sigma_k = 1^R a_jk k(a_ki = a_jk) to the principal axes form or cononical form (10), where lambda_1, ..., lambda_n are the (not necessarily distinct) eigenvalues of the (symmetric!) matrix A, and X is an orthogonal matrix with corresponding eigenvectors x_1, ..., x respectively, as column vectors. A quadratic form Q(x) = x^T Ax and its (symmetrical!) matrix A are called positive definite if Q(x) > 0 for all x notequal 0 negative definite if Q(x)

Solution

Answer of (a):

First assume that A be a real symmetric matrix of order n.

Since A is a real symmetric matrix, all it\'s eigen values are real. Let c₁,c₂......,c_n be the eigen values of A.

Since A is a real symmetric matrix, there exists an orthogonal matrix P such that P^-1AP (=P^t A P) is a diagonal matrix.But A and P^-1AP have the same eigen values.Therefore,

P^-1AP = diag(c₁,c₂......,c_n), where diag(c₁,c₂......,c_n) is the diagonal matrix whose diagonal elements are c₁,c₂......,c_n. Let A be positive definite. Then A is congruent to I_n, Where I_n is the identity matrices of order n. Therefore P^t A P is congruent to A .[since P is non-singular]

Consequently, P^tAP is congruent to I_n. Therefore c_i >0 for all i=1,2,.....,n.

Conversely,

               Let c_i >0 for all i=1,2,.....,n. Then diag(c₁,c₂......,c_n) is positive definite. But A is congruent to P^tAP = diag(c₁,c₂......,c_n). Therefore A is positive definite.

Answer of (b):

First assume that A be a real symmetric matrix of order n.

Since A is a real symmetric matrix, all it\'s eigen values are real. Let c₁,c₂......,c_n be the eigen values of A.

Since A is a real symmetric matrix, there exists an orthogonal matrix P such that P^-1AP (=P^t A P) is a diagonal matrix. But A and P^-1AP have the same eigen values.Therefore,

P^-1AP = diag(c₁,c₂......,c_n), where diag(c₁,c₂......,c_n) is the diagonal matrix whose diagonal elements are c₁,c₂......,c_n. Let A be negative definite. Then A is congruent to -I_n, Where I_n is the identity matrices of order n. Therefore P^t A P is congruent to A .[since P is non-singular]

Consequently, P^tAP is congruent to -I_n. Therefore c_i <0 for all i=1,2,.....,n.

Conversely,

               Let c_i <0 for all i=1,2,.....,n. Then diag(c₁,c₂......,c_n) is negative definite. But A is congruent to P^tAP = diag(c₁,c₂......,c_n). Therefore A is negatve definite.